What are the solutions to the equation (x-3)(x+4)(x+5)(x-4)=180 ?

The solutions to the equation (x-3)(x+4)(x+5)(x-4)=180 are x=1,x=-2,x=5,x=-6

Find the zeros of (x-3)(x+4)(x+5)(x-4)=180

The zeros of (x-3)(x+4)(x+5)(x-4)=180 are x=1,x=-2,x=5,x=-6

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(x - 3) (x + 4) (x + 5) (x - 4) = 180

x = 1, x = - 2, x = 5, x = - 6

Solution steps

(x - 3) (x + 4) (x + 5) (x - 4) = 180

Expand

(x - 3) (x + 4) (x + 5) (x - 4) : x^{4} + 2 x^{3} - 31 x^{2} - 32 x + 240

x^{4} + 2 x^{3} - 31 x^{2} - 32 x + 240 = 180

Move

180

to the left side

x^{4} + 2 x^{3} - 31 x^{2} - 32 x + 60 = 0

Factor

x^{4} + 2 x^{3} - 31 x^{2} - 32 x + 60 : (x - 1) (x + 2) (x - 5) (x + 6)

(x - 1) (x + 2) (x - 5) (x + 6) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x + 2 = 0 or x - 5 = 0 or x + 6 = 0

Solve

x - 1 = 0 : x = 1

Solve

x + 2 = 0 : x = - 2

Solve

x - 5 = 0 : x = 5

Solve

x + 6 = 0 : x = - 6

The solutions are

x = 1, x = - 2, x = 5, x = - 6

6 - 3 (x + 1) = - 4 x^{2} (1 - x)

3 x^{4} - 3 x^{3} - 1 x^{2} = 0

z \cdot z^{3} + 15 = 16.66

z^{3} - z^{2} + 4 z - 4 = 0

x^{4} - (4 \cdot x) + 5 = 0

What are the solutions to the equation (x-3)(x+4)(x+5)(x-4)=180 ?
The solutions to the equation (x-3)(x+4)(x+5)(x-4)=180 are x=1,x=-2,x=5,x=-6
Find the zeros of (x-3)(x+4)(x+5)(x-4)=180
The zeros of (x-3)(x+4)(x+5)(x-4)=180 are x=1,x=-2,x=5,x=-6